Question

What is the percent yield of AlBr3 if 131.6 grams of AlBr3 is produced in when 19.4 g of aluminum reacts? Round to two decimal places and do not include a % sign in your answer.

2Al + 3Br2 → 2AlBr3

Answer 1

19.4g Xg

2Al + 3Br₂ → 2AlBr₃

2*27g 2*267g

m_theoretical(AlBr₃) = X = 19.4 * 2*267 / (2*27) g = 191.84 g.

The percent yield of AlBr₃: η = m_practical(AlBr₃) / m_theoretical(AlBr₃) * 100% = 131.6 g / 191.84 g * 100% = 68.5973% ≈ 68.60%.

Answer: 68.60.